Question

If both the blocks as shown in the given arrangement are given together a horizontal velocis right. If a$_{cm}$ be the subsequent acceleration of the centre of mass of the system of block equals.

• A. 0 m/s²
• B. $\frac{5}{3}$ m/s²
• C. $\frac{7}{3}$ m/s²
• D. 2 m/s²

Answer 1

Answer: (D)

2 m/s²

Answer 2

To find the acceleration of the center of mass ($a_{cm}$) of the system, we need to consider the net external force acting on the system and the total mass of the system.

1. Identify the System and its Total Mass: The system consists of two blocks:

• Top block: $m_1 = 1$ kg
• Bottom block: $m_2 = 2$ kg The total mass of the system is $M = m_1 + m_2 = 1 \text{ kg} + 2 \text{ kg} = 3 \text{ kg}$.

2. Identify External Forces: The blocks are given a horizontal velocity to the right. The only external horizontal force acting on the system is the kinetic friction between the bottom block ($m_2$) and the ground. The friction between $m_1$ and $m_2$ is an internal force to the system, so it does not affect the acceleration of the center of mass of the entire system.

3. Calculate the Normal Force from the Ground: The total normal force exerted by the ground on the system is due to the combined weight of both blocks: $N_{total} = (m_1 + m_2)g$ Assuming $g = 10 \text{ m/s}^2$: $N_{total} = (1 \text{ kg} + 2 \text{ kg}) \times 10 \text{ m/s}^2 = 3 \text{ kg} \times 10 \text{ m/s}^2 = 30 \text{ N}$.

4. Calculate the External Kinetic Friction Force: The coefficient of kinetic friction between the 2 kg block and the ground is $\mu_2 = 0.2$. The kinetic friction force ($f_k$) acting on the system from the ground opposes the motion (acts to the left): $f_k = \mu_2 N_{total} = 0.2 \times 30 \text{ N} = 6 \text{ N}$. Since this force acts to the left, the net external force $F_{net, ext} = -6 \text{ N}$.

5. Calculate the Acceleration of the Center of Mass: The acceleration of the center of mass of a system is given by the net external force divided by the total mass: $a_{cm} = \frac{F_{net, ext}}{M}$ $a_{cm} = \frac{-6 \text{ N}}{3 \text{ kg}}$ $a_{cm} = -2 \text{ m/s}^2$

The magnitude of the acceleration of the center of mass is $2 \text{ m/s}^2$. The negative sign indicates that the acceleration is to the left, opposing the initial velocity.